E = row total × column total / grand total, df = (r − 1)(c − 1)
Critical values (α = 0.05)
df 1: 3.841, 2: 5.991, 3: 7.815, 4: 9.488
Decision
reject H0 if χ² > χ²crit
2Practice problems
1.Normal distributionMedium
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 98.6 s. Round z to two decimals.
μX = 60, σX = 5μY = 45, σY = 4X + Y > 98.6 s
P =
2.Normal distributionHard
The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.
μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
3.Confidence intervalsMedium
A sample of 11 items has mean 21.5 and standard deviation s = 7. The population is roughly normal. Find the lower limit of the 95% confidence interval for the mean.
n = 11x̄ = 21.5s = 795%t(10) = 2.228
lower limit =
4.Confidence intervalsHard
The population standard deviation is 26. How large a sample n do you need so that a 99% confidence interval for the mean has a margin of error of at most 0.5?
σ = 26E = 0.599%
n =
5.Hypothesis tests & confidence intervalsMedium
7 employees take a typing test before and after a training course. For each person the difference d = after − before (words per minute) is recorded. The mean difference is 3.62 and the standard deviation of the differences is 3.3. Find the paired test statistic t for H₀: μd = 0.
n = 7d̄ = 3.62sd = 3.3
t =
6.Hypothesis tests & confidence intervalsHard
A one-sided z test at α = 0.05 must detect a shift of the mean of 2.5 units with a power of 95%. The standard deviation is 15. Find the minimum sample size n.
The table gives 5 paired observations of x = temperature in °C and y = ice creams sold. Find the coefficient of determination r² of the least-squares line.
n = 5
r² =
8.Linear regression & correlationHard
The table gives 5 paired observations of x = load in kN and y = extension in mm. Use the least-squares line ŷ = a + b·x. A new observation (3, 14) is made. Find its residual e = y − ŷ.
n = 5(3, 14)
e =
9.Chi-square testsMedium
The contingency table shows observed counts. Compute the chi-square statistic χ² for the test of independence between the row and column variables. Do not use a continuity correction.
N = 802 × 2
χ² =
10.Chi-square testsHard
A chi-square goodness-of-fit test checks whether counts in 5 categories follow a Poisson distribution whose mean is estimated from the same data. How many degrees of freedom df does the test have?
k = 5
df =
3Solutions
1.Normal distribution
Answer: P = 0.8413
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the sum
μ = 105 s, σ = √(5² + 4²) = 6.403 s
Standardise
z = (98.6 − 105)/6.403 = −1
Upper tail
P = 1 − Φ(−1) = 1 − 0.1587 = 0.8413
Answer
P = 0.8413
2.Normal distribution
Answer: P = 0.8413
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Distribution of the difference
μ = 500 − 490 = 10 g, σ = 10 g
Standardise
z = (0 − 10)/10 = −1
Probability
P = 1 − Φ(−1) = 0.8413
Answer
P = 0.8413
3.Confidence intervals
Answer: lower limit = 16.8
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Critical value
df = n − 1 = 10, t = 2.228
Margin of error
E = 2.228·7/√11 = 4.702
Interval
21.5 ± 4.702 → [16.8, 26.2]
Answer
lower limit = 16.8
4.Confidence intervals
Answer: n = 17944
Confidence interval for a mean: x̄ ± margin of error. With a known population σ the margin is z·σ/√n; with only the sample standard deviation s, use t with n − 1 degrees of freedom: t·s/√n. Critical values: z = 1.645 (90%), 1.96 (95%), 2.576 (99%). To get a margin E or smaller, you need n ≥ (z·σ/E)², rounded up.
Solve for n
n = (2.576·26/0.5)² = 17943
Round up
A sample size is a whole number, and rounding down would make the margin too large.
n = 17944
Answer
n = 17944
5.Hypothesis tests & confidence intervals
Answer: t = 2.902
Testing a mean: the test statistic measures how many standard errors the sample mean lies from the value claimed in H₀: z = (x̄ − μ₀)/(σ/√n) when σ is known, t = (x̄ − μ₀)/(s/√n) when it is estimated from the sample. A confidence interval is x̄ ± (critical value)·(standard error).
Standard error of the mean difference
SE = 3.3/√7 = 1.247
Test statistic
t = 3.62/1.247 = 2.902
Answer
t = 2.902
With df = 6, the two-sided 5% critical value is 2.447; |t| is larger, so H₀ is rejected.
6.Hypothesis tests & confidence intervals
Answer: n = 390
Sample size for a test: to detect a true shift δ with power 1 − β in a one-sided z test at level α, the standard error must satisfy δ = (z_α + z_β)·σ/√n, so n = ((z_α + z_β)·σ/δ)², rounded up.
Solve for n
n = ((1.645 + 1.645)·15/2.5)² = 389.7
Round up
n = 390
Answer
n = 390
7.Linear regression & correlation
Answer: r² = 0.9904
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 120/5 = 24
ȳ = 547/5 = 109.4
Sums of squares and products
Sxx = Σ(x − x̄)² = 90
Sxy = Σ(x − x̄)(y − ȳ) = 537
Syy = Σ(y − ȳ)² = 3235
Coefficient of determination
r² = 537²/(90·3235) = 0.9904
Answer
r² = 0.9904
About 99.04% of the variation in y is explained by the line.
8.Linear regression & correlation
Answer: e = 3.8
Least-squares regression: with x̄ and ȳ the means, Sxx = Σ(x − x̄)², Syy = Σ(y − ȳ)² and Sxy = Σ(x − x̄)(y − ȳ), the line ŷ = a + b·x has slope b = Sxy/Sxx and intercept a = ȳ − b·x̄, so it always passes through (x̄, ȳ). Pearson's correlation coefficient r = Sxy/√(Sxx·Syy) lies between −1 and 1; r² is the fraction of the variation in y explained by the line.
Means
x̄ = 30/5 = 6
ȳ = 90/5 = 18
Sums of squares and products
Sxx = Σ(x − x̄)² = 40
Sxy = Σ(x − x̄)(y − ȳ) = 104
Regression line
b = 104/40 = 2.6
a = 18 − 2.6·6 = 2.4
ŷ = 2.4 + 2.6·x
Predicted value
ŷ = 2.4 + 2.6·3 = 10.2
Residual
e = 14 − 10.2 = 3.8
Answer
e = 3.8
The point lies above the line.
9.Chi-square tests
Answer: χ² = 2.143
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.
df = 1. At α = 0.05 the critical value is 3.841, so H₀ (independence) is not rejected.
10.Chi-square tests
Answer: df = 3
Chi-square tests: compare observed counts O with the counts E expected under H₀ using χ² = Σ(O − E)²/E. Goodness of fit: E = n·p for each category, df = k − 1 (minus one more for each parameter estimated from the data). Independence in an r × c table: E = row total × column total / grand total, df = (r − 1)(c − 1). Reject H₀ when χ² exceeds the critical value for the chosen α.